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This is my first post here. I apologize if it goes against any guidelines for posting. I study math as a hobby and am currently dealing with trigonometry on a high school level. I have so far learned the formulas for trigonometric addition and subtraction and double the angle, as well as what is in my language referred to as the ’trigonometric one’ - getting the radius of the unit circle by use of the pythagorean theorem. I have not yet gotten to deriving trigonometric functions. The following is a problem I could solve by plugging in a set of numbers, but in seeking a more elegant solution, perhaps, I found myself stuck and I don’t know what I am missing. I am appreciative of any help I get. The problem is as follows:

Show that if $A$ is an angle and $0^\circ<A<90^\circ$ then $\hspace{0.3cm}\left( 1+\dfrac {1}{\sin A}\right) \left( 1+\dfrac {1}{\cos A}\right)>5$

I began with the following assumption:

$$0^\circ<A<90^\circ\rightarrow0<{\sin A}<1\\0<{\cos A}<1\rightarrow\dfrac {1}{\sin A}\\\dfrac {1}{\cos A}>1$$

Given the above, it would follow that:

$$\begin{aligned} \lim _{A\rightarrow 90^\circ}\dfrac {1}{\cos A}&=\infty \\ \lim _{A\rightarrow 0^\circ}\dfrac {1}{\sin A}&=\infty \end{aligned}$$

This alone doesn’t seem like enough to show what is asked. I can show that at $A=45^\circ$ the product is still greater than 5, but I am not sure how any offset in degrees from there affects two trigonometric terms such that the product is still greater than 5. I also tried solving the inequality but ended up with fractioned terms I couldn’t add up or a cubic function if you will, that I couldn’t solve.

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    $\begingroup$ do you know trig identities such as $\sin^2 x+\cos^2 x=1$? $\endgroup$ – Vasya Jul 2 '18 at 15:42
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    $\begingroup$ Ignore $\lim_{A\to 90}$ We care about all values not just the the ones on the edges. $\endgroup$ – fleablood Jul 2 '18 at 15:48
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    $\begingroup$ You shouldn't use $\to$ to mean $\implies.$ They denote entirely different concepts. $\endgroup$ – Allawonder Jul 2 '18 at 22:08
  • $\begingroup$ Vasya, yes I do. I just didn’t know if it had a particular name. $\endgroup$ – Grenadine Jul 4 '18 at 15:50
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Hint:

Since $A$ is sharp you can write $\sin A = b/c$ and $\cos A = a/c$ where $c$ is hypotenuse in right triangle $ABC$ ($C=90$). Does that help?


Any way, since for all positive $x,y$ we have $x+y\geq 2\sqrt{xy}$: $$1+{1\over \sin A}\geq {2\over \sqrt{\sin A}}$$ and $$1+{1\over \cos A}\geq {2\over \sqrt{\cos A}}$$

so $$(1+{1\over \sin A})(1+{1\over \cos A})\geq {4\sqrt{2}\over \sqrt{\sin 2A}}\geq 4\sqrt{2} >5$$

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  • $\begingroup$ Thank you Angle! $\endgroup$ – Grenadine Jul 4 '18 at 15:52
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They probably expected something like this:

Expanding out, your inequality is the same as $$1 + {1 \over \sin A} + {1 \over \cos A} + {1 \over \sin A \cos A} \geq 5$$ This is equivalent to $${1 \over \sin A} + {1 \over \cos A} + {1 \over \sin A \cos A} \geq 4$$ Which is the same as $${1 \over \sin A} + {1 \over \cos A} + {2 \over \sin 2A} \geq 4$$ Since in the range in question, $0 < \sin A, \cos A < 1$ and $0 < \sin 2A \leq 1$, one has $${1 \over \sin A} + {1 \over \cos A} + {2 \over \sin 2A} > 1 + 1 + 2$$ $$= 4$$

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If we open the parenthesis, we get $$1+{1\over \sin A}+ {1\over \cos A}+{1\over \sin A \cos A}\geq 5$$ $${1+\sin A+ \cos A\over \sin A \cos A}\geq 4$$ $$1+\sin A+ \cos A -4 \sin A \cos A \ge 0$$ $$(\cos A - \sin A)^2+\cos A(1-\sin A)+\sin A(1-\cos A) \ge 0$$ In the last inequality, each term is not negative so the sum is not negative

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@Angle has given a simple way of solving the problem. If you know a little calculus, you could do it this way.

Let $$f(A)=\left( 1+\dfrac {1}{\sin A}\right) \left( 1+\dfrac {1}{\cos A}\right).$$ Then by the Quotient Rule, $$\small f'(A)=\left(0+\frac{0\sin A-1\cos A}{\sin^2A}\right)\left( 1+\dfrac {1}{\cos A}\right)+\left( 1+\dfrac {1}{\sin A}\right)\left( 0+\frac{0\cos A-1(-\sin A)}{\cos^2A}\right)$$ so $$\small f'(A)=-\frac{\cos A}{\sin^2A}\left( 1+\dfrac {1}{\cos A}\right)+\frac{\sin A}{\cos^2A}\left( 1+\dfrac {1}{\sin A}\right)=-\frac{\cos A}{\sin^2A}-\frac1{\sin^2A}+\frac{\sin A}{\cos^2A}+\frac1{\cos^2A}$$ giving $$f'(A)=\frac{1+\sin A}{\cos^2A}-\frac{1+\cos A}{\sin^2A}=0$$ for stationary points.

Thus $$\sin^2A(1+\sin A)=\cos^2A(1+\cos A)\implies \sin^2A-\cos^2A+\sin^3A-\cos^3A=0$$ so $$(\sin A-\cos A)(\sin A+\cos A)+(\sin A-\cos A)(\sin^2A+\sin A\cos A+\cos^2A)=0$$ giving $$(\sin A-\cos A)(1+\sin A+\cos A+\sin A\cos A)=0$$ and clearly one solution is when $\tan A=1\implies A=45^\circ$.

For the other equation, we solve $$1+\sin A+\cos A+\sin A\cos A=(1+\sin A)(1+\cos A)=0$$ but the solutions are outside of the range of $A$.

Hence $A=45^\circ$.

Now at this angle, $$f(45^\circ)=(1+\sqrt2)^2=3+2\sqrt2>3+2=5.$$

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    $\begingroup$ How do you deduce that $\sin^2 A + \sin^3 A = \cos^2 A + \cos^3 A$ implies that $\sin^2 A = \cos^2 A$ and $\sin^3 A = \cos^3 A$? When the latter two equalities are satisfied, the first clearly is, so we have a stationary point; but I don't see why (in this argument) there can be no other stationary points, which might have smaller values. $\endgroup$ – tomsmeding Jul 2 '18 at 20:23
  • $\begingroup$ @tomsmeding I agree. Please see the edit now. $\endgroup$ – TheSimpliFire Jul 3 '18 at 5:49
  • $\begingroup$ That looks great! $\endgroup$ – tomsmeding Jul 3 '18 at 7:58
  • $\begingroup$ Thanks, TheSimpleFire. I know some calculus but I haven’t gotten to things like the quotient rule yet. $\endgroup$ – Grenadine Jul 4 '18 at 15:49
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Use $AM-GM$ and $\sin^2 A + \cos^2 A = 1$.

.....

I start by simply doing.

$(1 + \frac {1}{\sin A})(1 + \frac 1{\cos A}) = $

$(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A}$.

Now $0 < \sin A < 1$ so $ \frac 1{\sin A} > 1$ and $0 < \cos A < 1$ so $ \frac 1{\cos A} > 1$.

So $(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A} > 3 + + \frac 1{\sin A \cos B}$.

Now common sense tells us $\sin A \cos A < 1*1$ so $\frac 1{\sin A \cos A} > 1$ and that gives us:

$(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A}) > 4$ and that is not good enough.

So I need more. Now I know $\sin A$ and $\cos A$ aren't just any numbers less than one. I know $\sin^2 A + \cos^2 A = 1$ and I know that if gets close to $1$ or $0$ the other gets close to $0$ and $1$ and I know that $\sin 45 = \cos 45 = \frac{\sqrt {2}}{2}$ but other wise one is more and the other is less than $\frac{\sqrt{2}}{2}$.

This all triggers that I can probably use the AM-GM theorem to "boost" $\frac 1 {\sin A\cos b} > 1$ to $\frac 1{\sin A\cos B} > 2$. Maybe... I'll use AM-GM and see what happens.

So we need to use AM-GM to prove $\sin A \cos B < \frac 12$ and $\frac 1{\sin A \cos A}> 2$.

Now AM- GM states for $M,N > 0$ thatn $\frac {M+N}2 \ge \sqrt{MN}$ so $\sin A\cos A = \sqrt{\sin^2 A\cos^2 A} \le \frac {\sin^2 A + \cos^2 A}2 = \frac 12$.

ANd that's that.

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Another way $\cos A = \sin (90 - A)$. Let $A = 45 + b$ then

$\sin A = \sin (45+b) = \sin 45 \cos b + \cos 45 \sin b=\frac {\sqrt 2}2(\cos b + \sin b)$ and $\cos A = \sin (45 - b) = \sin 45 \cos b - \cos 45 \sin b=\frac {\sqrt 2}2(\cos b - \sin b)$.

And $\sin A*\cos A = (\frac {\sqrt 2}2)^2(\cos b + \sin b)(\cos b - \sin b) = \frac 12(\cos^2 b - \sin^2 b)= \frac 12(1-2\sin^2 b) < \frac 12$

So $(1 +\frac 1{\sin A})(1 + \frac 1 {\cos A}) =$

$1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1 {\sin A\cos B} > 1 + 1+1+2 =5$.

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    $\begingroup$ What is $B$, and why can you write $\sin^2 A + \cos^2 B = 1$? $\endgroup$ – aschepler Jul 3 '18 at 0:30
  • $\begingroup$ B is a typo for A. $\endgroup$ – fleablood Jul 3 '18 at 5:10
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Let $y=\dfrac{1+\sin A+\cos A+\sin A\cos A}{\sin A\cos A}$

$\iff(y-1)\sin A\cos A-1=\sin A+\cos A$

Squaring both sides we get $$1+\sin2A=1+\dfrac{(y-1)^2\sin^22A}4-(y-1)\sin2A$$

$\iff(y-1)^2\sin^22A-4y\sin2A=0$

As $\sin2A>0,$ $$(y-1)^2\sin2A-4y=0$$

$\implies4y=(y-1)^2\sin2A\le(y-1)^2$

$\iff y^2-6y+1\ge0\implies y>\dfrac{6+\sqrt{32}}2=3+2\sqrt2$ as $y>0$

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